Verified Solution • 2025–26 Session moderate

QJEEM25Maf779: JEE Main 2025 Mathematics - Class 12 JEEM25Maf779 Solution: Let , where is the constant of integration. If , then equals :

Let ∫x3sin⁡x dx=g(x)+C\int x^3 \sin x \mathrm{~d} x=g(x)+C , where CC is the constant of integration. If 8(g(π2)+g′(π2))=απ3+βπ2+γ,α,β,γ∈Z8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z , then α+β−γ\alpha+\beta-\gamma equals :

Multiple Choice Options

  • A 47
  • B 55
  • C 62
  • D 48

Step-by-Step Verified Solution

amp;∫x3sin⁡xdx=−x3cos⁡x+∫3x2cos⁡xdxamp;=−x3cos⁡x+3x2sin⁡x−∫6xsin⁡xdxamp;=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x+c\begin{aligned}& \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \\& =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \\& =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c\end{aligned}

So g(x)=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡xg(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x

amp;g(π2)=3π24−6amp;g′(x)=x3sin⁡xamp;g′(π2)=π38amp;8(g(π2)+g′(π2))=π3+6π2−48\begin{aligned}& g\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \\& g^{\prime}(x)=x^3 \sin x \\& g^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \\& 8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48\end{aligned}

So α+β−γ=55\alpha+\beta-\gamma=55

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